# Why doesn't two topologies being homeomorphic imply that the induced map between the topologies is a homeomorphism?

Suppose that $h : X \rightarrow Y$ is a homeomorphism. The homeomorphism $h$ induces a bijection $h_* : \tau_{\small X} \rightarrow \tau_{ \small Y}$ between the topologies of $X$ and $Y$.

My question is: what do we mean when we say that the two topologies $\tau_{\small X}$ and $\tau_{\small Y}$ are homeomorphic? I understand that a homeomorphism $h$ between $X$ and $Y$ means that $X$ and $Y$ are homeomorphic as sets/spaces, but when we say that $\tau_{\small X}$ and $\tau_{\small Y}$ are homeomorphic, aren't we just saying that there is a homeomorphism between some subsets of $\mathcal{P}(X)$ and $\mathcal{P}(Y)$, respectively, which would mean that $h_*$ is not only a bijection, but also a homeomorphism itself? Or have I just got confused between the notion/definition of homeomorphic topologies?

Edit: Also, as I side question, I'm guessing that $h_*$ is the thing which maps open sets to open sets, not necessarily $h$. Is there an example of a homeomorphism $h$ which doesn't map open sets to open sets?

Edit 2: (from the comments) If we have a homeomorphism map $h : X \rightarrow X$, and we give a topology $\tau_1$ to the domain, and a different topology $\tau_2$ to the codomain, is it possible that the domain and codomain won't be homeomorphic?

Many thanks for any answers.

• I don't recall anyone ever saying that two topologies were homeomorphic. Where have you seen/heard this usage? – John Hughes Dec 28 '17 at 19:14
• In an exercise I was doing a while ago, though it's possible I misunderstood the wording. – mathphys Dec 28 '17 at 19:17
• If we have a homeomorphism $h: X \rightarrow X$, and we give a topology $\tau_1$ to the domain, and a different topology $\tau_2$ to the codomain, is it possible that the domain and codomain won't be homeomorphic? I think that this may be the source of my confusion. – mathphys Dec 28 '17 at 19:21
• A homeomorphism is something between spaces. A space is a set with a topology. Using notation from your comment, say $X$ is a set. Then what you mean when you say $h:X\to X$ is a homeomorphism is either $h:(X,\tau_1)\to(X,\tau_2)$ in which case, yes it is a homeomorphism because you said it is. Or it is $h:(X,\tau_1)\to(X,\tau_1)$ is a homeomorphism and if we take the same set map but with a different topology of the codomain $h’:(X,\tau_1)\to(X,\tau_2)$ is that a homomorphism too? The answer to that is not necessarily. Take $X=\mathbb R$, $\tau_1$ Euclidean, and $\tau_2$ discrete, $h=id.$ – Dan Robertson Dec 28 '17 at 19:31
• A question you may be interested in is related to the fact that the open sets of a topology make up a complete lattice and a homeomorphism induces a lattice isomorphism. One may ask when a lattice isomorphism between lattices of open sets is induced by a homeomorphism. – Robert Thingum Dec 28 '17 at 19:54

1. No, there is no such an example. If $h$ is a homeomorphism and $A$ is an open subset of $X$, then $h^{-1}$ is continuous, and therefore $(h^{-1})^{-1}(A)$ is an open subset of $Y$. But $(h^{-1})^{-1}(A)=h(A)$.
2. Yes, even if the map is continuous. For instance, take $X=\mathbb R$, with the discrete topology on the left and the usual one on the right.
The answer to question $2$ is emphatically yes. Let $\tau_1$ be the indiscrete topology (only $X$ and $\emptyset$ open) and let $\tau_2$ be the discrete topology (every set open); it's easy to check that if $X$ has at least two elements, the identity map is not a homeomorphism (and indeed there is no homeomorphism).
Note, though, that the way you've phrased that question is a bit odd: the particular map $h$ doesn't come into play at all. I think what you meant was, "if $h$ is a map $X\rightarrow X$ and $\tau_1, \tau_2$ are topologies on $X$, is $h$ necessarily a homeomorphism from $(X, \tau_1)$ to $(X, \tau_2)$?" But note that this isn't quite what you asked.